Question 14

Based on the principles of qualitative analysis for basic radicals:

• (P) Group IV reagent: $H_2S$ in the presence of $NH_4OH$ precipitates $Mn^{2+}$ as $MnS$. So, P $\rightarrow$ 3.
• (Q) Group V reagent: $(NH_4)_2CO_3$ in the presence of $NH_4OH$ precipitates $Ba^{2+}$ as $BaCO_3$. So, Q $\rightarrow$ 4.
• (R) Group III reagent: $NH_4OH$ in the presence of $NH_4Cl$ precipitates $Al^{3+}$ as $Al(OH)_3$. So, R $\rightarrow$ 2.
• (S) Group II reagent: $H_2S$ in the presence of dilute $HCl$ precipitates $Cu^{2+}$ as $CuS$. So, S $\rightarrow$ 1.
• $Mg^{2+}$ (5) belongs to Group VI.

The matching sequence is P-3, Q-4, R-2, S-1. Correct Option: A